Problem 3.17. Torsion of open and closed section cantilever

Rotation of the closed section beam, ψ=

Rotation of the open section beam, ψ=

Step 1. Calculate the torsional stiffnesses of cross sections a) and b)

Check torsional stiffness of the closed sectionHide torsional stiffness

Eq.(3-111)

$G{I}_{t,\mathrm{closed}}=\frac{4A_c^2}{\oint {\displaystyle \frac{1}{Gt}}ds}=\frac{4{\left(h–t\right)}^4}{4\left(h–t\right)}Gt={\left(200–8\right)}^3\times 8\times 80.8\times {10}^3=4.575\times {10}^{12 }{\mathrm{Nmm}}^2=4575 {\mathrm{kNm}}^2$

Check torsional stiffness of the open sectionHide torsional stiffness

Eq.(3-100)

$G{I}_{t,\mathrm{open}}=\underset{k=1}{\overset{4}{G\sum }}\frac{b_k{t}_k^3}{3}=G\frac{4\left(h–t\right)t^3}{3}=80.8\times {10}^3\frac{4\left(200–8\right)8^3}{3}=1.059\times {10}^{10}{\mathrm{Nmm}}^2=10.59 {\mathrm{kNm}}^2$

Step 2. Draw the torque diagram and the rate of twist distribution along the length of the beam.

Check diagramsHide diagrams

Step 3. Determine the rotation of the beam end.

Check rotationHide rotation

Rotation is obtained by the integration of the rate of twist function along the beam’s length. The rotation of the beam end results in the area of the rate of twist diagram.

See geometrical equation in Table 3.7.

$\vartheta =\frac{d\psi }{dx} \to \psi ={\int }_0^L\vartheta dx=\frac{\vartheta L}{2}=\frac{t{L}^2}{2G{I}_t}$

for closed section beam

${\mathit{\psi }}_{\mathbf{closed}}=\frac{t{L}^2}{2G{I}_{t,\mathrm{closed}}}=\frac{4\times 2.5^2}{2\times 4575}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00273}$

for open section beam

${\mathit{\psi }}_{\mathbf{open}}=\frac{t{L}^2}{2G{I}_{t,\mathrm{open}}}=\frac{4\times 2.5^2}{2\times 10.59}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{18}$

The torsional stiffnesses of the closed cross section is

Eq.(3-111)

The torsional stiffnesses of the open cross section is

Eq.(3-100)

From uniformly distributed torque load the internal force diagram and the rate of twist distribution is linear along the length of the beam:

Rotation is obtained by the integration of the rate of twist function along the beam’s length. The rotation of the beam end results in the area of the rate of twist diagram.

See geometrical equation in Table 3.7.

$\vartheta =\frac{d\psi }{dx} \to \psi ={\int }_0^L\vartheta dx=\frac{\vartheta L}{2}=\frac{t{L}^2}{2G{I}_t}$

for closed section beam

${\mathit{\psi }}_{\mathbf{closed}}=\frac{t{L}^2}{2G{I}_{t,\mathrm{closed}}}=\frac{4\times 2.5^2}{2\times 4575}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00273}$

for open section beam

${\mathit{\psi }}_{\mathbf{open}}=\frac{t{L}^2}{2G{I}_{t,\mathrm{open}}}=\frac{4\times 2.5^2}{2\times 10.59}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{18}$